reserve r,p,x for Real;
reserve n for Element of NAT;
reserve A for non empty closed_interval Subset of REAL;
reserve Z for open Subset of REAL;
reserve a,b,x for Real;
reserve n for Element of NAT;
reserve A for non empty closed_interval Subset of REAL;
reserve f,f1,f2 for PartFunc of REAL,REAL;
reserve Z for open Subset of REAL;

theorem
  A c= Z & (for x st x in Z holds f.x=a-x & f.x>0) & dom ((-2)(#)(( #R (
1/2))*f)) = Z & dom ((-2)(#)(( #R (1/2))*f)) = dom f2 & (for x st x in Z holds
f2.x = (a-x) #R (-1/2)) & f2|A is continuous implies integral(f2,A) = ((-2)(#)(
  ( #R (1/2))*f)).(upper_bound A) -((-2)(#)(( #R (1/2))*f)).(lower_bound A)
proof
  assume that
A1: A c= Z and
A2: for x st x in Z holds f.x=a-x & f.x>0 and
A3: dom ((-2)(#)(( #R (1/2))*f)) = Z and
A4: dom ((-2)(#)(( #R (1/2))*f)) = dom f2 and
A5: for x st x in Z holds f2.x = (a-x) #R (-1/2) and
A6: f2|A is continuous;
A7: f2 is_integrable_on A by A1,A3,A4,A6,INTEGRA5:11;
A8: (-2)(#)(( #R (1/2))*f) is_differentiable_on Z by A2,A3,FDIFF_4:31;
A9: for x being Element of REAL
st x in dom (((-2)(#)(( #R (1/2))*f))`|Z) holds (((-2)(#)(( #R (1/2
  ))*f))`|Z).x = f2.x
  proof
    let x be Element of REAL;
    assume x in dom (((-2)(#)(( #R (1/2))*f))`|Z);
    then
A10: x in Z by A8,FDIFF_1:def 7;
    then (((-2)(#)(( #R (1/2))*f))`|Z).x = (a-x) #R (-1/2) by A2,A3,FDIFF_4:31
      .= f2.x by A5,A10;
    hence thesis;
  end;
  dom (((-2)(#)(( #R (1/2))*f))`|Z) = dom f2 by A3,A4,A8,FDIFF_1:def 7;
  then (((-2)(#)(( #R (1/2))*f))`|Z) = f2 by A9,PARTFUN1:5;
  hence thesis by A1,A3,A4,A6,A7,A8,INTEGRA5:10,13;
end;
